Question

Suppose there are 5 pairs of shoes in a bag and four shoes are taken out at random. What is the probability that among the four taken out there is at least one complete pair?

Answer 1

there are 5 pairs of shoe
s = {5 }
let event A be 4 Pairs of shoes are remove randomly
A = {4}
p(A) = n(A)
n(S)
p(A) = 4/5


therefore p(A) = 4/5

Answer 2

The probability that among the four taken out there is at least one complete pair is $\frac{20}{21}$.

Given: 5 pairs of shoes are in a bag and four shoes are taken out at random.

To Find: The probability that among the four taken out there is at least one complete pair.

Solution: Let P = no pair of shoes were chosen

A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the selection order does not matter.

$^n C_r=\frac{n !}{r ! (n-r) !}$

where $^n C_r =$ the number of combinations

n = total number of objects in the set

r = number of choosing objects from the set.

⇒ We have to select four power to take 1 shoe from each one of them out of 5 = $^{5}C_{4}$

and, 2 choices for each = $^{5}C_{4}$ × 2

And, total choice = $^{10}C_{4}$

⇒ P = $\frac{2(^{5}C_{4})}{^{10}C_{4}}$ = $\frac{10}{^{10}C_{4}}$

Required probability = 1 - P = 1 - $\frac{10}{^{10}C_{4}}$

= 1 - 10 × 24

= 1 - $\frac{1}{21}$

= $\frac{20}{21}$

So, the probability that among the four taken out there is at least one complete pair is $\frac{20}{21}$.

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